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こんにちは、php コードに問題があります。採用システム用に 3 つのテーブルがあります。

CREATE TABLE IF NOT EXISTS `members` (
   `uid` int(11) NOT NULL AUTO_INCREMENT,
   `username` varchar(25) CHARACTER SET utf8 COLLATE utf8_bin NOT NULL,
   `password` varchar(25) CHARACTER SET utf8 COLLATE utf8_bin NOT NULL,
   `cpassword` varchar(25) CHARACTER SET utf8 COLLATE utf8_bin NOT NULL,
   `email` varchar(30) CHARACTER SET utf8 COLLATE utf8_bin NOT NULL,
   `role` varchar(30) CHARACTER SET utf8 COLLATE utf8_bin NOT NULL,
   PRIMARY KEY (`uid`)'
) ENGINE=InnoDB  DEFAULT CHARSET=latin1 AUTO_INCREMENT=60 ;

候補表:

CREATE TABLE IF NOT EXISTS `candidate` (
  `fullname` varchar(35) CHARACTER SET utf8 COLLATE utf8_bin NOT NULL,
  `webpage` varchar(150) CHARACTER SET utf8 COLLATE utf8_bin NOT NULL,
  `tel` int(35) NOT NULL,
  `nationality` varchar(35) CHARACTER SET utf8 COLLATE utf8_bin NOT NULL,
  `position` varchar(30) CHARACTER SET utf8 COLLATE utf8_bin NOT NULL,
  `interviewed` varchar(30) NOT NULL DEFAULT 'No',
  `rating` varchar(30) NOT NULL,
  `c_id` int(11) NOT NULL AUTO_INCREMENT,
  `uid` int(11) NOT NULL,
  PRIMARY KEY (`c_id`),
  KEY `uid` (`uid`)
) ENGINE=InnoDB  DEFAULT CHARSET=latin1 AUTO_INCREMENT=135 ;

アカデミック テーブル : 

CREATE TABLE IF NOT EXISTS `academic_candidate` (    
    `degree` varchar(30) CHARACTER SET utf8 COLLATE utf8_bin NOT NULL,
    `exp_years` varchar(25) CHARACTER SET utf8 COLLATE utf8_bin NOT NULL,
    `comment1` varchar(300) CHARACTER SET utf8 COLLATE utf8_bin NOT NULL,
    `proposed_positions` varchar(25) CHARACTER SET utf8 COLLATE utf8_bin NOT NULL,
    `research_years` varchar(25) CHARACTER SET utf8 COLLATE utf8_bin NOT NULL,
    `comment2` varchar(300) CHARACTER SET utf8 COLLATE utf8_bin NOT NULL,
    `department` varchar(25) CHARACTER SET utf8 COLLATE utf8_bin NOT NULL,
    `a_id` int(25) NOT NULL AUTO_INCREMENT,
    `uid` int(11) NOT NULL,
    `c_id` int(11) NOT NULL,
    PRIMARY KEY (`a_id`),
    UNIQUE KEY `id` (`a_id`),
    KEY `uid` (`uid`),
    KEY `c_d` (`c_id`),
    KEY `c_id` (`c_id`)
) ENGINE=InnoDB  DEFAULT CHARSET=latin1 AUTO_INCREMENT=40 ;`

-- テーブルの制約academic_candidate

ALTER TABLEcademic_candidatecademic_candidate_ibfk_1 uidメンバーのuidcademic_candidate_ibfk_2c_id候補者c_id ADD CONSTRAINT_FOREIGN KEY () REFERENCES() ON DELETE CASCADE ON UPDATE CASCADE, ADD CONSTRAINTFOREIGN KEY () REFERENCES() ON DELETE CASCADE ON UPDATE CASCADE;

--

-- テーブルの制約candidate

ALTER TABLE候補の ADD CONSTRAINTcandidate_ibfk_1 FOREIGN KEY (uid) REFERENCESメンバーの(uid) ON DELETE CASCADE ON UPDATE CASCADE;

--

今、私はこのクエリを使用して、テーブルのacademic_candidateに値を保存します

session_start();

    $query = "SELECT * FROM academic_candidate WHERE degree = '$degree'";
$result = mysql_query($query);
$count  = mysql_num_rows($result);

if($count > 0){ 
    echo "You ALready complete the form </br>";
    header("Location:../candidate/candidate_index.php");
}
else{
        $degree =($_POST['degree']);
        $exp_years = ($_POST['exp_years']);
        $comment1 = ($_POST['comment1']);
        $proposed_positions = ($_POST['proposed_positions']);
    $research_years=($_POST['research_years']);
    $comment2=($_POST['comment2']);
    $department=($_POST['department']);

   $uid=($_SESSION['uid']);


    $query1 = "INSERT INTO academic_candidate

                (degree,exp_years,comment1,proposed_positions,research_years,comment2,department,uid,c_id)
                SELECT
                 '$degree','$exp_years','$comment1','$proposed_positions','$research_years','$comment2','$department','$uid','$c_id'
                    FROM  candidate         
                 WHERE  uid='$uid' AND c_id='$c_id'  ";


    $result = mysql_query($query1);

    if(!$result){
        echo "Error";
        die (mysql_error());                                          
    }   

    else{
        header("Location:../candidate/view_application.php");
    }

}

私の問題は、テーブルcademic_candidateテーブルにすべての値を格納することですが、 c_id は 0 です。Candidate.c_id を取得するにはどうすればよいですか?

4

1 に答える 1

0

INSERT INTOリテラル値をステートメントに渡しています。

違いを考えてみましょう:

SELECT 'bar' FROM foo;

SELECT bar FROM foo;

これは SQLFiddle のデモです。

あなたのクエリを見る：

SELECT '$degree'

次と同じではありません:

SELECT degree

これらの 1 つは、渡された値を単純にエコーすることであり、もう 1 つは実際に DB から列を選択することです。

于 2013-08-13T17:29:37.563 に答える